Type inference in TypeScript with transitivity

Consider this code snippet for illustration:

function foo(t: "number"): number
        function foo(t: "string"): string
        function foo(t: "boolean"): boolean
        function foo(t: "number" | "string" | "boolean"): number | string | boolean {
            if (Math.random() < 0.5) {
                if (t === "number") {
                    return 1;
                } else if (t === "boolean") {
                    return true;
                } else {
                    return "1";
                }
            } else {
                return foo(t);
            }
        }

The error message is as follows:

No overload matches this call.
      Overload 1 of 3, '(t: "number"): number', gave the following error.
        Argument of type '"string" | "number" | "boolean"' is not assignable to parameter of type '"number"'.
          Type '"string"' is not assignable to type '"number"'.
      Overload 2 of 3, '(t: "string"): string', gave the following error.
        Argument of type '"string" | "number" | "boolean"' is not assignable to parameter of type '"string"'.
          Type '"number"' is not assignable to type '"string"'.
      Overload 3 of 3, '(t: "boolean"): boolean', gave the following error.
        Argument of type '"string" | "number" | "boolean"' is not assignable to parameter of type '"boolean"'.
          Type '"string"' is not assignable to type '"boolean"'.(2769)

On the other hand, TypeScript successfully checks the following code:

function bar(t: "number"): number
        function bar(t: "string"): string
        function bar(t: "boolean"): boolean
        function bar(t: "number" | "string" | "boolean"): number | string | boolean {
            if (Math.random() < 0.5) {
                if (t === "number") {
                    return 1;
                } else if (t === "boolean") {
                    return true;
                } else {
                    return "1";
                }
            } else {
                switch (t) {
                    case "number": return bar(t);
                    case "string": return bar(t);
                    case "boolean": return bar(t);
                }
            }
        }

Refer to TypeScript Playground.

I am curious why TypeScript disapproves of transitive type inference in the initial code snippet. Any insights on this matter would be greatly appreciated. Thank you.

javascripttypescript

Answer №1

I may not have the solution to "how", but here is a suggestion:

function foo<T extends "number" | "string" | "boolean">(t: T): T {

If you require something more complex, there's an inconveniently lengthy option:

function foo<T extends "number" | "string" | "boolean">(t: T): T extends "number" ? number : T extends "string" ? string : boolean

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